Root Space Decomposition

2.5 Semisimple Lie Algebras

2.5.2 Root Space Decomposition

g=^M

λ

g_λ(X),

andSacts byλong_λ(X). TakingD=adX,Y∈g_λ(X),Z∈g_µ(X), andksufficiently large in (2.45), we see that

[g_λ(X),g_µ(X)]⊂g_λ+µ(X). (2.46) HenceSis a derivation ofg. By Corollary 2.5.12 there existsXs∈gsuch that adXs=

S. SetX_n=X−X_s. ut

2.5 Semisimple Lie Algebras 115

XY v=Y X v+ [X,Y]v=0

for allX∈h, since[X,Y]∈h. ThusW is invariant underY, so there exists a nonzero vectorv0∈W such thatY v0=0. It follows thatgv₀=0. ut Corollary 2.5.15.There exists a basis for V in which the elements ofgare repre- sented by strictly upper-triangular matrices.

Proof. This follows by repeated application of Theorem 2.5.14, replacingV by

V/Cv₀at each step. ut

Corollary 2.5.16.Suppose g is a semisimple Lie algebra. Then there exists a nonzero element X∈gsuch thatadX is semisimple.

Proof. We argue by contradiction. Ifgcontained no nonzero elementsX with adX semisimple, then Corollary 2.5.13 would imply that adX is nilpotent for allX∈g.

Hence Corollary 2.5.15 would furnish a basis forgsuch that adX is strictly upper triangular. But then adXadY would also be strictly upper triangular for allX,Y∈g, and hence the Killing form would be zero, contradicting Theorem 2.5.11. ut For the rest of this section we letgbe a semisimple Lie algebra. We call a subal- gebrah⊂gatoral subalgebraif adXis semisimple for allX∈h. Corollary 2.5.16 implies the existence of nonzero toral subalgebras.

Lemma 2.5.17.Lethbe a toral subalgebra. Then[h,h] =0.

Proof. LetX∈h. Thenhis an invariant subspace for the semisimple transformation adX. If[X,h]6=0 then there would exist an eigenvalueλ 6=0 and an eigenvector Y ∈hsuch that[X,Y] =λY. But then

(adY)(X) =−λY6=0, (adY)²(X) =0,

which would imply that adY is not a semisimple transformation. Hence we must

have[X,h] =0 for allX∈h. ut

We shall call a toral subalgebrah⊂gaCartan subalgebraif it has maximal di- mension among all toral subalgebras ofg. From Corollary 2.5.16 and Lemma 2.5.17 we see thatgcontains nonzero Cartan subalgebras and that Cartan subalgebras are abelian. We fix a choice of a Cartan subalgebrah. Forλ∈h^∗let

g_λ={Y∈g:[X,Y] =hλ,XiY for all X∈h}.

In particular,g₀={Y ∈g: [X,Y] =0 for all X∈h}is thecentralizerofhing.

LetΦ ⊂g^∗\ {0}be the set ofλ such thatg_λ 6=0. We callΦ the set ofrootsofh ong. Since the mutually commuting linear transformations adXare semisimple (for X∈h), there is aroot space decomposition

g=g₀⊕^M

λ∈Φ

g_λ .

LetBdenote the Killing form ofg. By the same arguments used for the classical groups in Sections 2.4.1 and 2.4.2 (but now usingBinstead of the trace form on the defining representation of a classical group), it follows that

1. [g_λ,g_µ]⊂g_λ+µ ;

2. B(g_λ,g_µ) =0 ifλ+µ6=0 ;

3. the restriction ofBtog₀×g₀is nondegenerate;

4. ifλ∈Φthen−λ∈Φand the restriction ofBtog_λ×g_−λ is nondegenerate.

New arguments are needed to prove the following key result:

Proposition 2.5.18.A Cartan algebra is its own centralizer ing; thush=g₀. Proof. Sincehis abelian, we haveh⊂g₀. LetX∈g₀and letX=X_s+X_nbe the Jordan decomposition ofXgiven by Corollary 2.5.13.

(i) X_sandX_nare ing₀.

Indeed, since [X,h] =0 and the adjoint representation of g is faithful, we have [Xs,h] =0. HenceX_s∈hby the maximality ofh, which implies thatX_n=X−X_sis also inh.

(ii) The restriction ofBtoh×his nondegenerate.

To prove this, let 06=h∈h. Then by property (3) there exists X ∈g₀ such that B(h,X)6=0. SinceX_n∈g₀by (i), we have[h,X_n] =0 and hence adhadX_nis nilpotent ong. ThusB(h,X_n) =0 and soB(h,X_s)6=0. SinceX_s∈h, this proves (ii).

(iii) [g0,g₀] =0 .

For the proof of (iii), we observe that ifX∈g₀, then adX_sacts by zero ong₀, since X_s∈h. Hence adX|^g0=adX_n|^g0 is nilpotent. Suppose for the sake of contradiction that [g0,g₀]6=0 and consider the adjoint action of g₀ on the invariant subspace [g0,g₀]. By Theorem 2.5.14 there would exist 06=Z∈[g0,g₀]such that[g0,Z] =0.

Then[g₀,Z_n] =0 and hence adYadZ_nis nilpotent for allY ∈g₀. This implies that B(Y,Z_n) =0 for allY ∈g₀, so we conclude from (3) thatZ_n=0. ThusZ=Z_smust be inh. Now

B(h,[X,Y]) =B([h,X],Y) =0 for all h∈h and X,Y ∈g₀. Henceh∩[g0,g₀] =0 by (ii), and soZ=0, giving a contradiction.

It is now easy to complete the proof of the proposition. IfX,Y∈g₀then adX_nadY is nilpotent, sinceg₀is abelian by (iii). HenceB(Xn,Y) =0, and soX_n=0 by (3).

ThusX=X_s∈h. ut

Corollary 2.5.19.Let gbe a semisimple Lie algebra andha Cartan subalgebra.

Then

g=h⊕^M

λ∈Φ

g_λ. (2.47)

Hence if Y ∈gand [Y,h]⊂h, then Y ∈h. In particular,his a maximal abelian subalgebra ofg.

2.5 Semisimple Lie Algebras 117

Since the formBis nondegenerate onh×h, it defines a bilinear form onh^∗that we denote by(α,β).

Theorem 2.5.20.The roots and root spaces satisfy the following properties:

1.Φ spansh^∗.

2. Ifα ∈Φ thendim[gα,g_−α] =1and there is a unique element hα ∈[gα,g_−α] such thathα,hαi=2(call hα thecoroottoα).

3. Ifα∈Φ and c∈Cthen cα∈Φif and only if c=±1. Alsodimg_α=1.

4. Letα,β∈Φ withβ6=±α. Let p be the largest integer j≥0withβ+jα∈Φ and let q be the largest integer k≥0withβ−kα∈Φ. Then

hβ,hαi=q−p∈Z (2.48)

andβ+rα∈Φ for all integers r with−q≤r≤p. Henceβ− hβ,h_αiα∈Φ. 5. Ifα,β ∈Φ andα+β∈Φ, then[g_α,gβ] =g_α+β .

Proof. (1): Ifh∈handhα,hi=0 for allα∈Φ, then[h,g_α] =0 and hence[h,g] =0.

The center ofgis trivial, sinceghas no abelian ideals, soh=0. ThusΦspansh^∗. (2): LetX∈g_αandY∈g_−α. Then[X,Y]∈g₀=hand forh∈hwe have

B(h,[X,Y]) =B([h,X],Y) =hα,hiB(X,Y).

Thus[X,Y]corresponds toB(X,Y)α under the isomorphismh∼=h^∗given by the formB. SinceBis nondegenerate ong_α×g_−α, it follows that dim[gα,g_−α] =1.

SupposeB(X,Y)6=0 and setH= [X,Y]. Then 06=H∈h. Ifhα,Hi=0 thenH would commute withXandY, and hence adHwould be nilpotent by Lemma 2.5.1, which is a contradiction. Hencehα,Hi 6=0 and we can rescaleX andY to obtain elementse_α∈g_α andf_α∈g_−α such thathα,h_αi=2, whereh_α= [eα,f_α].

(3): Lets(α) =Span{eα,fα,hα} ∼=sl(2,C)and set Mα=Chα+

∑

c6=0

g_cα.

Since[eα,g_cα]⊂g_(c+1)α, [fα,g_cα]⊂g_(c−1)α, and[eα,g_−α] = [fα,g_α] =Ch_α, we see thatMα is invariant under the adjoint action ofs(α).

The eigenvalues of adh_αonMα are 2cwith multiplicity dimg_cαand 0 with mul- tiplicity one. By the complete reducibility of representations ofsl(2,C)(Theorem 2.3.6) and the classification of irreducible representations (Proposition 2.3.3) these eigenvalues must be integers. Hencecα∈Φimplies that 2cis an integer. The eigen- values in any irreducible representation are all even or all odd. Hencecα is not a root for any integercwith|c|>1, sinces(α)contains the zero eigenspace inM_α. This also proves that the only irreducible component ofM_α with even eigenvalues iss(α), and it occurs with multiplicity one.

Suppose(p+1/2)α ∈Φ for some positive integer p. Then adh_α would have eigenvalues 2p+1,2p−1, . . . ,3,1 onM_α, and hence(1/2)α would be a root. But

thenα could not be a root, by the argument just given, which is a contradiction.

Thus we conclude that Mα=Chα+Ceα+Cfα. Hence dimg_α=1.

(4): The notion ofα root string throughβ from Section 2.4.2 carries over verba- tim, as does Lemma 2.4.3. Hence the argument in Corollary 2.4.5 applies.

(5): This follows from the same argument as Corollary 2.4.4. ut